Air Resistance in Orbit

The International Space Station (ISS) isn’t exactly in space. Well, it depends on what you mean by “space”. Is there gravity? Of course. Is there air? Not very much, but there is a tiny bit. This tiny bit of air does push on the space station, just a little bit. If left alone, the ISS would eventually slow down and move to a lower orbit where there is even more air. Yes, it would eventually crash.

But fear not. The ISS can get a reboots from the Automated Transfer Vehicle (ATV). Basically, the ATV uses it’s fuel to accelerate the ISS back up to it’s proper orbit. And that is what you see in the video above – the inside view showing the acceleration of the ISS.

Is it possible to measure the acceleration of the ISS based on the motion of the astronauts? Could you use video analysis to do this? Yes and yes. Do you want to see the details? Head over the ESA blog and check out my guest post. Besides just finding the acceleration, I talk about fake forces in accelerating reference frames. I also estimate the acceleration due to the air resistance based on the acceleration and frequency of the reboost.

You know I can’t just give you a link to something else. I have to add something to this. So here we go.

An Estimation For the Density of Air in Orbit

In my calculation of the reboost acceleration, I estimated an average drag force on the ISS with a value of about 0.9 Newtons. (Note: I just realized that this value of force is based on a reboost time around 2 min. Actual reboost times are around 30-40 minutes – so this number is off. I will let you fix the values below for a homework assignment) Now for some wild speculation. Suppose that I can use the typical model for air resistance that has the following form:

This air resistance model works pretty well for things moving at sub sonic speeds on the surface of the Earth. The ISS clearly is not in this same category. However, we can still use it to get a rough estimate. In this model, ρ is the density of the air, A is the cross sectional area of the object and C is some drag coefficient that depends on the shape of the object.

If I have an estimate for the magnitude of this air resistance force and the speed of the ISS, I can find the product of ρ A and C. Let’s say the ISS has an orbital speed of 7.7 x 103 m/s (as listed on the Wikipedia page on the ISS). This would put the product of unknowns at:

Now for some wild estimates about A and C. Wikipedia lists some drag coefficients for different shapes. Something streamlined would have a low coefficient – like 0.04 where a smooth brick will have a value around 2.0. What about about the ISS? It clearly is not streamlined. How about an estimate from 1.5 – 2.0?

What about the cross sectional area? This is a tough one. If the solar panels are perpendicular to the direction of motion, the area will be much higher than if it is not. The specifications list a width of 108 m with a height of 20 m. Of course all of this space is not filled in. Let me estimate a cross section area from 200 m2 to 800 m2.

This puts the product of A and C anywhere from 300 to 1600 m2. Instead of picking one value to calculate the density, I will make a graph of density vs. values of AC.

What is the density of air in orbit? From this, I get a density around 10-10 to 10-11 kg/m3. That’s pretty low – as would be expected.